The constant ‘c’ of Rolle’s theorem for the function f(x)=sinx in [0,2π] is

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π6

π3

π4

π2

answer is D.

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Detailed Solution

f1(x)=cosx and f(0)=f(2π)=0 c∈(0,2π) such that f'(c)=0⇒cosc=0⇒c=π2 or 3π2

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